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The total magnification produced by a co...

The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14cm.
If the least distance of distinct vision is 20cm, calculate the focal length of the objective and the eye piece.

Text Solution

Verified by Experts

Given , M = 20, `m_(e ) = 5, D = 20cm`.
`m_(e ) = ( D )/( u_(e )) ` ( For eyepiece ) ,br> `5 = ( 20) /( u_(e ))`
`u _(e ) = 4 cm`
`u _(e ) = - 4cm, v _(e) =D= - 20cm`
`(1)/(f_(e)) = (1)/( v_(e)) - (1)/( u_(e ))`
`= ( 1)/( -20) +(1)/( 4)`
`= ( -1+5)/(20) = ( 4)/( 20)`
Hence, focal length of eyepiece,
`f_(e ) = 5 cm`
`M = m_(0) xx m_(e )`
`m_(0) = ( M)/( m_(e )) = ( 20)/( 5) = 4`
`v_(0) = 14-4 =10cm`
For objective lens,
`M_(0) = - ( v_(0))/( u_(0))`
`u= - (10)/( u ) ( :. M_(0)=4)`
`rArr u_(0) = - (10)/( u ) = - 2.5 cm`
Now, `(1)/(f_(0)) = ( 1)/(v_(0)) - ( 1)/(u_(0))`
`= ( 1)/( 10) - (10)/( - 2.5)`
`= (1)/(10) + (10)/( 2.5)`
`= (1)/(10) + ( 10)/( 25)`
`= (1)/(2)`
`:.` `f_(0) = 2 cm`
Hence, focal length of objective `f_(0) = 2.0 cm`.
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